Respuesta :

check the picture below.

so it hits the ground when y = 0, thus

[tex]\bf ~~~~~~\textit{initial velocity}\\\\ \begin{array}{llll} ~~~~~~\textit{in feet}\\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{16}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{5}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases}[/tex]

[tex]\bf h(t)=-16t^2+16t+5\implies \stackrel{h(t)}{0}=-16t^2+16t+5 \\\\\\ 16t^2-16t-5=0\implies (4t+1)(4t-5)=0\\\\ -------------------------------\\\\ 4t+1=0\implies 4t=-1\implies t=-\cfrac{1}{4}\\\\ -------------------------------\\\\ 4t-5=0\implies 4t=5\implies t=\cfrac{5}{4}\implies \boxed{t=1\frac{1}{4}}[/tex]

since "t" is seconds it took, it can't be a negative amount.
Ver imagen jdoe0001
ashishdwivedilVT

Time taken by the rock to hit the ground is [tex]t=\frac{8+\sqrt{89} }{5}[/tex].

given,

initial velocity u = 16 ft/s

initial height h' = 5 feet

time of flight= ????....let us say 't'

gravitational acceleration=10 m/s²

angle of projection is not given, let us say it is 90degree

what is the projectile motion?

It is the motion of an object thrown up in the air at an angle from the horizontal (or ground), an object moving under gravitational acceleration pointed vertically downwards.

the object when projected will reach up to the maximum height, and then it will come back to the ground due to the effect of gravity.

as we know that empirical relation between height, initial velocity, and time of flight is

[tex]h(t) = usin90*t-\frac{1}{2} gt^{2} + h'\\[/tex]

the time when a rock will just hit the ground the height will be zero.

we can say that

0=  16t-1/2*10*t²+5

5t²-16t-5=0

[tex]t=\frac{8+\sqrt{89} }{5}[/tex] m/s

time taken by a rock to hit the ground is [tex]t=\frac{8+\sqrt{89} }{5}[/tex].

to get more about projectile motion refer to the link,

https://brainly.com/question/24216590

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